mongodb - 如何在php中进行mongo db聚合连接操作？

Question

我在 mongodb 中有三个集合，例如

    employeeTitleTbl contains 

    {
    "_id": ObjectId("5a7c2027d2ccda04d0006070"),
    "EmployeeTitleName": "Librarian",
    "EmployeeTitleCode": "ji8766",
   ..
    }

 DepartmentsTbl contains

  {
   "_id": ObjectId("5a817402d2ccda13240015a5"),
   "DepartmentCode": "149",
   "DepartmentName": "Biology",
  ..
 }

 EmployeesTbl contains 

  {
   "_id": ObjectId("5a81957bd2ccda13240015a8"),
    "EmployeeNumber": "25789998",
   "JobTitle": ObjectId("5a7c2008d2ccda04d000606f"),
   "Department": ObjectId("5a8173e6d2ccda13240015a4"),
       ...
  }

我对mongodb很新，而且驱动也更新了3.6，所以很难在php中找到最新的聚合方法。所以我正在调整下面的代码，以便在EmployeesTbl中添加部门和员工职位的实际值，例如“EmployeeTitleName”和“DepartmentName”。

  public function fetchAll()
{         
        $pipeline = array(
                            array(
                                '$lookup' => array(
                                'from' => 'DepartmentsTbl',
                                'localField' => 'Department',
                                'foreignField' => '_id',
                                'as' => 'DepartmentDetails'
                                      )
                                  ),

                                array(
                                '$lookup' => array(
                                'from' => 'employeeTitleTbl',
                                'localField' => 'JobTitle',
                                'foreignField' => '_id',
                                'as' => 'JobTitleDetails'
                                      )
                                  ),
                         );

                 try
                 {                    

                    $cursor = $this->db->EmployeesTbl->aggregate($pipeline);

                 }

                 catch(Exception $e){

                 }

    return $cursor->toArray();
}           

请用最新的方法帮助我！！！

输出就像

array(2) { [0]=> object(MongoDB\Model\BSONDocument)#29 (1) { ["storage":"ArrayObject":private]=> array(25) { ["_id"]=> object (MongoDB\BSON\ObjectId)#10 (1) { ["oid"]=> string(24) "5a8176c0d2ccda13240015a6" } ["EmployeeNumber"]=> string(8) "25799989".... ... ....... ["DepartmentDetails"]=> object(MongoDB\Model\BSONArray)#27 (1) { ["storage":"ArrayObject":private]=> array(0) { } } [" JobTitleDetails"]=> object(MongoDB\Model\BSONArray)#28 (1) { ["storage":"ArrayObject":private]=> array(0) { } } } [1]=..... ……

Answer 1

例如，您需要使用 $lookup 来连接 2 个或更多表。

 db.employeeTitle.aggregate([{
  $lookup: {
    from: 'employees',
    localField: 'EmployeeTitleName',
    foreignField: '_id',
    as: 'employeeData'
  }
 },
 { $unwind: '$employeeData' },
 {
   $project: { profile: '$employeeData' }     
 }    
])

它连接两个表employeeTitle 和employees，并为您提供有关员工的完整信息。